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Current Electricity — JEE physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceJEE physicsCurrent Electricity1 Mark
MCQ
An ionization chamber with parallel conducting plates as anode and cathode has $5 \times 10^7$ electrons and the same number of singlycharged positive ions per $\mathrm{cm}^3$. The electrons are moving at $0.4 \mathrm{~m} / \mathrm{s}$. The current density from anode to cathode is $4 \mu \mathrm{A} / \mathrm{m}^2$. The velocity of positive ions moving towards cathode is
  • A
    $0.4 \mathrm{~m} / \mathrm{s}$
  • B
    $16 \mathrm{~m} / \mathrm{s}$
  • C
    Zero
  • ✓
    $0.1 \mathrm{~m} / \mathrm{s}$
✓

Answer

Correct option: D.
$0.1 \mathrm{~m} / \mathrm{s}$
Current density of drifting electrons $j=n e v$
$ n=5 \times 10^7 \mathrm{~cm}^{-3}=5 \times 10^7 \times 10^6 \mathrm{~m}^{-3} .$
$ v=0.4 \mathrm{~ms}^{-1}, e=1.6 \times 10^{-19} \mathrm{C} \Rightarrow j=3.2 \times 10^{-6} \mathrm{Am}^{-2}$
Current density of ions $=(4-3.2) \times 10^{\circ}=0.8 \times 10^{-6} \frac{A}{\mathrm{~m}^2}$
This gives $v$ for ions $=0.1 m s$.

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